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- Question 1 of 10
1. Question
Average age of a family (father, mother, son and daughter) is 32.5 years. If son got married and the average becomes decreased by 0.9 years, then find the daughter-in-law’s age
CorrectSum of the ages of a family (Exclude daughter-in-law) =130 years
Sum of the ages of a family (Include daughter-in-law) =158 years
Daughter-in-law’s age= 158-130=28 yearsIncorrectSum of the ages of a family (Exclude daughter-in-law) =130 years
Sum of the ages of a family (Include daughter-in-law) =158 years
Daughter-in-law’s age= 158-130=28 years - Question 2 of 10
2. Question
A, B and C enters into a partnership with a capital ratio of 4:5:6. After 4 months, A withdraws half of the capital and B and C invest additionally 20% and 40% of their previous capital respectively. What is the profit ratio at the end of a year?
CorrectProfit ratio of A, B and C = (4x*4+2x*8) : (5x*4+5x*120/100*8): (6x*4+6x*140/100*8)
=>320:680:912
=>40:85:114IncorrectProfit ratio of A, B and C = (4x*4+2x*8) : (5x*4+5x*120/100*8): (6x*4+6x*140/100*8)
=>320:680:912
=>40:85:114 - Question 3 of 10
3. Question
HCF and LCM of two numbers are in the ratio of 1:66 and the difference of the two numbers is 35. Find the HCF of the two numbers if the ratio of the two numbers is 6:11
CorrectTwo numbers are 42 and 77
HCF*LCM= Product of two numbers
X*66x=42*77
X2=49
X=7
IncorrectTwo numbers are 42 and 77
HCF*LCM= Product of two numbers
X*66x=42*77
X2=49
X=7
- Question 4 of 10
4. Question
Three pipes can fill an empty tank in 4 hours. If there is leakage in the tank it takes 1 hour more, then find the no of hours required to empty the full tank by leakage pipe alone
CorrectRequired no of hours = ¼ -1/5 =(5-4)/20=1/20
IncorrectRequired no of hours = ¼ -1/5 =(5-4)/20=1/20
- Question 5 of 10
5. Question
A bucket contains some quantity of milk and water, in the ratio of water and milk is 3:5. 40 litres of mixture is drawn and replaced with water and the ratio of milk and water becomes 5:11 then find the initial quantity of milk
Correct(5x-25)/(3x-15+40)=5/11
55x-275=15x+125
=>x=10 litres
Initial quantity of milk = 10*5=50 litresIncorrect(5x-25)/(3x-15+40)=5/11
55x-275=15x+125
=>x=10 litres
Initial quantity of milk = 10*5=50 litres - Question 6 of 10
6. Question
A bag contains x+4 pink, 6 green and 8 brown colour balls; if two balls are taken random and the probability of getting both are green colour balls is 5/92, then find the difference between the no. of pink colour balls and the no. of brown colour balls
CorrectGiven,
6c2/(x+18)C2=5/92
X2+35x-246=0
Simplify the above equation we get x=6
Required difference = 10-8=2 balls
IncorrectGiven,
6c2/(x+18)C2=5/92
X2+35x-246=0
Simplify the above equation we get x=6
Required difference = 10-8=2 balls
- Question 7 of 10
7. Question
A and B enter into a partnership with the capitals are in the ratio of 7:8. After some months C joins with them x% of B’s capital and received two-seventh of total profit. At the end of a year total profit Rs. 6400, in how many months C in the business
CorrectThe given data is insufficient to answer the given question.
IncorrectThe given data is insufficient to answer the given question.
- Question 8 of 10
8. Question
A and B together can do a piece of work in 6 2/3 days. B and C together can do the same work in 8 4/7 days. C and D together can do the same work in 12 days. All of them working together then find the total wage if C received Rs.180 per day
CorrectThe given data is insufficient to answer the given question
IncorrectThe given data is insufficient to answer the given question
- Question 9 of 10
9. Question
The average monthly salary of 25 employees in an organization is Rs. 3000. If the manager’s salary is added, then the average salary increases by Rs. 500. What is the manager’s monthly salary?
CorrectManager’s monthly salary = Rs. (3500 × 26 – 3000 × 25) = Rs. 16000.
IncorrectManager’s monthly salary = Rs. (3500 × 26 – 3000 × 25) = Rs. 16000.
- Question 10 of 10
10. Question
The sum of the present ages of a father and his son is 75 years. Five years ago father’s age was four times the age of the son. After 4 years, son’s age will be:
CorrectLet the present ages of son and father be x and (75 – x) years respectively.
Then, (75 – x) – 5 = 4 (x – 5) ⇔ 70 – x = 4x – 20 ⇔ 5x = 90 ⇔ x = 18.
∴ Son’s age after 4 years = (x + 4) = 22 years.IncorrectLet the present ages of son and father be x and (75 – x) years respectively.
Then, (75 – x) – 5 = 4 (x – 5) ⇔ 70 – x = 4x – 20 ⇔ 5x = 90 ⇔ x = 18.
∴ Son’s age after 4 years = (x + 4) = 22 years.
